Chapter 13, Problem 10P

(a)

To determine

The radius of the $\text{C}$ nucleus.

Answer to Problem 10P

The radius of the $\text{C}$ nucleus is $2.75\times {10}^{-15}\text{ m}$ .

Explanation of Solution

Write the equation for the nuclear radius.

  $R=R_0{A}^{1/3}$        (I)

Here, $R$ is the nuclear radius, $R_0$ is constant with value $1.2\times {10}^{-15}\text{ m}$ and $A$ is the mass number.

Conclusion:

The mass number of $\text{C}$ is $12$ .

Substitute $1.2\times {10}^{-15}\text{ m}$ for $R_0$ and $12$ for $A$ in equation (I) to find $R$ .

  $\begin{array}{c}R=\left(1.2\times {10}^{-15}\text{ m}\right){\left(12\right)}^{1/3}\\ =2.75\times {10}^{-15}\text{ m}\end{array}$

Therefore, the radius of the $\text{C}$ nucleus is $2.75\times {10}^{-15}\text{ m}$ .

(b)

To determine

The force of repulsion between a proton at the surface of a $\text{C}$ nucleus and the remaining five protons.

Answer to Problem 10P

The force of repulsion between a proton at the surface of a $\text{C}$ nucleus and the remaining five protons is $152\text{ N}$ .

Explanation of Solution

Write the equation for the Coulombic force.

  $F=\frac{k{q}_1{q}_2}{r^2}$

Here, $F$ is the force, $k$ is the Coulomb constant, $q_1,q_2$ are the charges and $r$ is the distance between the charges.

The charge of proton is $+e$ and the charge o the remaining protons is $\left(Z-1\right)e$ , where $Z$ is the atomic number. The value of $r$ will be equal to $R$ .

Substitute $+e$ for $q_1$ , $\left(Z-1\right)e$ for $q_2$ and $R$ for $r$ in the above equation.

  $F=\frac{k\left(Z-1\right)e^2}{R^2}$        (II)

Here, $e$ is the magnitude of the electronic charge.

Conclusion:

The value of $k$ is $9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ and the charge of the electron is $1.60\times {10}^{-19}\text{ C}$ . The atomic number of $\text{C}$ is $6$ .

Substitute $9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ for $k$ , $6$ for $Z$ , $1.60\times {10}^{-19}\text{ C}$ for $e$ and $2.75\times {10}^{-15}\text{ m}$ for $R$ in equation (II) to find $F$ .

  $\begin{array}{c}F=\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(6-1\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{{\left(2.75\times {10}^{-15}\text{ m}\right)}^2}\\ =152\text{ N}\end{array}$

Therefore, the force of repulsion between a proton at the surface of a $\text{C}$ nucleus and the remaining five protons is $152\text{ N}$ .

(c)

To determine

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

Answer to Problem 10P

The work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is $2.62\text{ MeV}$ .

Explanation of Solution

The work that must be done will be equal to the electrostatic potential energy of the system.

Write the equation for the electrostatic potential energy.

  $U=\frac{k{q}_1{q}_2}{r}$

Here, $U$ is the potential energy.

Substitute $+e$ for $q_1$ , $\left(Z-1\right)e$ for $q_2$ and $R$ for $r$ in the above equation.

  $U=\frac{k\left(Z-1\right)e^2}{R}$        (III)

Conclusion:

Substitute $9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ for $k$ , $6$ for $Z$ , $1.60\times {10}^{-19}\text{ C}$ for $e$ and $2.75\times {10}^{-15}\text{ m}$ for $R$ in equation (III) to find $U$ .

  $\begin{array}{c}U=\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(6-1\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{\left(2.75\times {10}^{-15}\text{ m}\right)}\\ =4.19\times {10}^{-13}\text{ J}\frac{1\text{ eV}}{1.60\times {10}^{-19}\text{ J}}\frac{1\text{ MeV}}{{10}^6\text{ eV}}\\ =2.62\text{ MeV}\end{array}$

Therefore, the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is $2.62\text{ MeV}$ .

(d)

To determine

The radius of the $\text{U}$ nucleus, force of repulsion between a proton at the surface of a $\text{U}$ nucleus and the remaining protons and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus.

Answer to Problem 10P

The radius of the $\text{U}$ nucleus is $7.44\times {10}^{-15}\text{ m}$ , the force of repulsion between a proton at the surface of a  nucleus $\text{U}$ and the remaining protons is $379\text{ N}$ and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is $17.6\text{ MeV}$ .

Explanation of Solution

The mass number of $\text{U}$ is $238$ and the atomic number is $92$ .

Conclusion:

Substitute $1.2\times {10}^{-15}\text{ m}$ for $R_0$ and $238$ for $A$ in equation (I) to find $R$ .

  $\begin{array}{c}R=\left(1.2\times {10}^{-15}\text{ m}\right){\left(238\right)}^{1/3}\\ =7.44\times {10}^{-15}\text{ m}\end{array}$

Substitute $9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ for $k$ , $92$ for $Z$ , $1.60\times {10}^{-19}\text{ C}$ for $e$ and $7.44\times {10}^{-15}\text{ m}$ for $R$ in equation (II) to find $F$ .

  $\begin{array}{c}F=\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(92-1\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{{\left(7.44\times {10}^{-15}\text{ m}\right)}^2}\\ =379\text{ N}\end{array}$

Substitute $9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2$ for $k$ , $92$ for $Z$ , $1.60\times {10}^{-19}\text{ C}$ for $e$ and $7.44\times {10}^{-15}\text{ m}$ for $R$ in equation (III) to find $U$ .

  $\begin{array}{c}U=\frac{\left(9\times {10}^9\text{ N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(92-1\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{\left(7.44\times {10}^{-15}\text{ m}\right)}\\ =2.82\times {10}^{-12}\text{ J}\frac{1\text{ eV}}{1.60\times {10}^{-19}\text{ J}}\frac{1\text{ MeV}}{{10}^6\text{ eV}}\\ =17.6\text{ MeV}\end{array}$

Therefore, the radius of the $\text{U}$ nucleus is $7.44\times {10}^{-15}\text{ m}$ , the force of repulsion between a proton at the surface of a nucleus $\text{U}$ and the remaining protons is $379\text{ N}$ and the work that must be done to overcome the electrostatic repulsion and put the last proton into the nucleus is $17.6\text{ MeV}$ .